- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 6. Hypothesis Testing
- Subject 8. Tests Concerning a Single Mean

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**CFA Practice Question**

In the past, the average length pregnancy in northwest Florida has been 270 days with a standard deviation of 16 days. A hospital in the area wants to show that the average length of pregnancy has been reduced. A random sample of 64 pregnancies yields a sample mean of 266 days. At an a-level of 5%, the hospital's conclusion should be ______.

B. There is no evidence that the average length of pregnancy in northwest Florida is more than in the past.

C. The average length of pregnancy in northwest Florida is more than in the past.

D. There is no evidence that the average length of pregnancy in northwest Florida is less than in the past.

A. The average length of pregnancy in northwest Florida is less than in the past.

B. There is no evidence that the average length of pregnancy in northwest Florida is more than in the past.

C. The average length of pregnancy in northwest Florida is more than in the past.

D. There is no evidence that the average length of pregnancy in northwest Florida is less than in the past.

Correct Answer: A

The null hypothesis is H

_{0}: μ ≥ 270. The alternative hypothesis is H_{a}: μ < 270. The critical value at an a-level of 10% is -1.645. The test value is (266 - 270)/(16/sqrt(64)) = -2. So, the decision is to reject H_{0}and the conclusion is that the average length of pregnancy in northwest Florida is less than in the past.###
**User Contributed Comments**
10

User |
Comment |
---|---|

GeoffT |
Where is the t table? |

laurenduvall |
why do we use a t table here? we know the standard deviation of the population. should we not use a z table? |

lawrence |
yes z-table is used in the explanation. 10% - 1.645 |

kaliokale |
but when u check the z table: 5%= -1.645 but not 10% |

mtcfa |
This is a one-tailed test: at the 5% significance level, the critical value is 1.645. Ths is parallel to the 2-sided test at 10% level of significance, where there is 5% IN EACH TAIL. Hope this helps solve some of the confusion. |

bobert |
266-270/ (16/ sqrt(64)) = -2 you SHOULD know the left tail with 5% is -1.96 standard deviations. There is no concern for the right tail, we want to know below. 2 > 1.96 so the p-val is less than 5% therefore it is a stat that falls in the distribution. You wont need to memorize the t and z tables, they will be provided if necessary, otherwise know the main confidence intervals and just compare. I recommend memorizing MINIMALLY these CI's: 99% 95% 90% 80% 68% |

cslau83 |
why can't null hypothesis be: mean <= 270? |

NikolaZ |
@cslau83: It says that they want to show it is less than 270 days. Thus, to show this they have to reject a null hypothesis in favor of the alternative, so the alternative must be mean < 270. If the null hypothesis was mean < 270, or mean =< 270, it wouldn't help them in proving anything - since of course failing to reject the null hypothesis does not imply that it is true. Remember - the null hypothesis can only be rejected or failed to be rejected (it cannot be proved). Thus, it is always the alternative hypothesis that we are proving. |

bidisha |
thanks nikola |

jgoff508 |
The alternative hypothesis, Ha, must contain the figures we're trying to prove or show. Many of these problems contain the phrases "...trying to prove..." or "...want(s) to show..." we use the figures in H(a). For example, in this problem: A hospital in the area WANTS TO SHOW (prove) that the average length of pregnancies have reduced. So, we use what is trying to be shown (proven) as the alternative hypothesis. We're given what the average has been historically. Now we're trying to prove that length of pregnancies now < historical average of 270. |